Question

If \[\left| z-3+2i \right|\le 4\] then the difference between the greatest value and the least value of \[\left| z \right|\] is
A. \[2\sqrt{113}\]
B. \[2\sqrt{13}\]
C. \[8\]
D. \[4+\sqrt{13}\]

Answer 1

Hint: First we will compare the given complex equation with the equation of the circle that is in complex form and find the center and the radius of the circle. Then we draw the normal of the PQ through the origin ‘O’ that passes through the center ‘C’ as because the least and the greatest distance always occurs along the normal. Later by using the distance formula we will get to know the least distance and the greatest distance and then the difference of these two will produce the required answer.

Formula used:
The equation of the circle in complex form is given by;
\[\left| z-{{z}_{0}} \right|=r\], where \[{{z}_{0}}\] the center of the circle and ‘r’ is the radius.

The distance between the two points using distance formula is,
\[d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\] Where \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\] are the coordinates of the two given points.

Complete step by step solution:
We have given that,
\[\left| z-3+2i \right|\le 4\]
The equation of the circle in complex form is given by;
\[\left| z-{{z}_{0}} \right|=r\], where \[{{z}_{0}}\] the center of the circle and ‘r’ is the radius.
Rewritten the given equation in the form of circle equation in complex form,
\[\left| z-\left( 3-2i \right) \right|\le 4\]
Therefore,
This represents an equation of the circle having the radius is 4 units and the center of the circle is \[\left( 3,-2 \right)\].
As we know that,
The normal always passes through the center of the circle. The greatest and the least distances always occur along the normal through the origin that passes through the center of the circle.
Now,
Let the normal be the PQ through the origin ‘O’ that passes through the center ‘C’.
The point of the centre ‘C’ having the coordinate \[\left( 3,-2 \right)\].
The given figure is as follows:

Hence,
We can see that;
\[OP\] is the least or the smallest distance; \[OP=CP-OC\].
And,
\[OQ\] is the largest or the greatest distance; \[OQ=CQ+OC\]
As we know that,
Radius = \[4\]
Thus,
\[CP=CQ=radius=4\]
As we know that,
The distance between the two points using distance formula is,
\[d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\] Where \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\] are the coordinates of the two given points.
Therefore,
\[OC=\sqrt{{{\left( 3-0 \right)}^{2}}+{{\left( -2-0 \right)}^{2}}}=\sqrt{9+4}=\sqrt{13}\]
We know that,
\[OP=CP-OC=4-\sqrt{13}\]
Therefore,
The least distance = \[OP=4-\sqrt{13}\]
Now,
We know that,
\[OQ=CQ+OC=4+\sqrt{13}\]
Therefore,
The greatest distance = \[OQ=4+\sqrt{13}\]
Thus,
The distance between the greatest and the least distance is given by;
\[\Rightarrow OQ-OP\]
\[\Rightarrow \left( 4+\sqrt{13} \right)-\left( 4-\sqrt{13} \right)\]
\[\Rightarrow 4+\sqrt{13}-4+\sqrt{13}\]
\[\Rightarrow 2\sqrt{13}\]
Hence,
The distance between the greatest and the least distance is \[2\sqrt{13}\].

Hence, the option (B) is the correct answer.

Note: While solving these types of questions, students need to remember the equation of the circle in a complex form i.e. \[\left| z-{{z}_{0}} \right|=r\], where \[{{z}_{0}}\] the center of the circle and ‘r’ is the radius. Students should know how to form an equation in complex form, while comparing any equation with \[\left| z-{{z}_{0}} \right|=r\] take care of the sign of \[{{z}_{0}}\]. For example; we have taken \[\left| z+1 \right|\] as \[\left| z-\left( -1 \right) \right|\]. Students should also know that the greatest and the least distances always occur along the normal through the origin that passes through the center of the circle.